<p>The coefficient of \(x^{n-2}\) in the polynomial \((x + 1)(x + 2)(x + 3)\cdots(x + n)\) is</p>
<p>(a) \(\frac{n(n^2 + 2)}{24}\)</p>
<p>(b) \(\frac{n^2(n + 1)(3n + 2)}{24}\)</p>
<p>(c) \(\frac{n(n + 1)(3n - 2)}{24}\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: The coefficient of $x^{n-2}$ in the expansion comes from selecting exactly $(n-2)$ factors of $x$ from the product, which means selecting the constant terms from exactly 2 factors. We need to find the sum of all products of pairs of constants from $\{1, 2, 3, \ldots, n\}$.
<p><strong>Step 1:</strong> Expand $(x+1)(x+2)(x+3)\cdots(x+n)$ using the multinomial theorem. The general term is obtained by selecting either $x$ or the constant from each factor.</p><p><strong>Step 2:</strong> For the coefficient of $x^{n-2}$, we need exactly $n-2$ factors to contribute $x$ and exactly $2$ factors to contribute their constants. This gives us:</p><p>Coefficient of $x^{n-2}$ = (sum of products of all pairs of constants from $\{1,2,3,\ldots,n\}$)</p><p><strong>Step 3:</strong> We need to find $e_2 = \sum_{1 \leq i < j \leq n} ij$.</p><p><strong>Step 4:</strong> Use the identity: $(1+2+3+\cdots+n)^2 = \sum_{i=1}^n i^2 + 2\sum_{1 \leq i<j \leq n} ij$</p><p>Therefore: $2e_2 = \left(\frac{n(n+1)}{2}\right)^2 - \sum_{i=1}^n i^2$</p><p><strong>Step 5:</strong> Calculate $\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$</p><p><strong>Step 6:</strong> $2e_2 = \frac{n^2(n+1)^2}{4} - \frac{n(n+1)(2n+1)}{6}$</p><p>$= \frac{n(n+1)}{12}[3n(n+1) - 2(2n+1)]$</p><p>$= \frac{n(n+1)}{12}[3n^2 + 3n - 4n - 2]$</p><p>$= \frac{n(n+1)}{12}[3n^2 - n - 2]$</p><p>$= \frac{n(n+1)(3n - 2)(n+1)}{12}$</p><p><strong>Step 7:</strong> Therefore: $e_2 = \frac{n(n+1)(3n-2)}{24}$</p><p><strong>∴ Answer:</strong> C</p>
Correct Answer: C